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I'm reading "Contemporary Abstract Algebra," by Gallian.

This is Exercise 4.51.$^\dagger$

Suppose that $G$ is a finite group with the property that every nonidentity element has prime order (e.g., $D_3$). If $Z(G)$ is not trivial, prove that every nonidentity element of $G$ has the same order.


Lemma: If $G$ is abelian with the property that every nonidentity element has prime order, then every nonidentity element has the same prime order.

Proof: If $G$ is abelian (i.e., $G=Z(G)$), then consider $g,h\in G$ such that $\lvert g\rvert=p$ and $\lvert h\rvert=q$ for distinct primes $p$ and $q$. We have

\begin{align} (gh)^{pq}&=(g^p)^q(h^q)^p\\ &=e, \end{align}

so that $\lvert gh\rvert$ divides $pq$.

If $\lvert gh\rvert=pq$, then it is composite, a contradiction; thus without loss of generality $\lvert gh\rvert=p$. Now we have

\begin{align} e&=(gh)^p\\ &=g^ph^p\\ &=eh^p \\ &=h^p, \end{align}

but now $q\mid p$, which is a contradiction since $p\neq q$ and $p$ is prime.

Thus all nonidentity elements of $G$ have the same prime order.$\square$

That's all I have so far.

I've considered proving some version of the contrapositive but nothing springs to mind other than, "yeah . . . contrapositive might work" followed by a shrug.


This comment gives me some idea of how to finish; however, I'm not sure where the finiteness of $G$ comes into play.

Please help :)

$\dagger$ I've just noticed that this exercise has a solution in the book. It makes sense to me. If anyone would like to answer it here anyway, go ahead! I might post an answer later summarising the proof in the text.

  • 2
    $\begingroup$ How about mimicking the abelian proof with one general element of the group and one nontrivial element of the center? $\endgroup$ – Mindlack Jan 16 at 11:55
  • $\begingroup$ I see what you mean, @Mindlack! That's a fun little trick! Thank you. $\endgroup$ – Shaun Jan 16 at 11:57
  • $\begingroup$ But . . . Why is finiteness necessary, then, in the original exercise? @Mindlack. $\endgroup$ – Shaun Jan 16 at 11:58
  • 3
    $\begingroup$ @Shaun They might just be trying to avoid awkward questions like "is infinity prime". $\endgroup$ – user3482749 Jan 16 at 12:17
  • 1
    $\begingroup$ @Shaun They don't. The problem is that it's kind of an awkward edge-case definition thing, and I wouldn't be surprised if someone just ruled out the infinite case to avoid having to think about it. $\endgroup$ – user3482749 Jan 16 at 12:24

The following paraphrases the proof in the solutions section of the book the exercise is from.

Let $z\in Z(G)$ such that $\lvert z\rvert=p$ for a prime $p$. Consider $g\in G$. We have $\lvert g\rvert=q$ is prime. Then

\begin{align} (zg)^{pq}&=(z^p)^q(g^q)^p \\ &=e^qe^p \\ &=e \end{align}

and thus $\lvert zg\rvert\in\{p, q\}$ (since it has to be prime). Assume without loss of generality that $\lvert zg\rvert=p$. Then

\begin{align} e&=(zg)^p \\ &=z^pg^p \\ &=eg^p \\ &=g^p, \end{align}

so $q\mid p$. Hence $p=q$.


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